Q376
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
The velocity of particle, when it travels 26 m is
A.
$21 \, ms^{-1}$
B.
$15 \, ms^{-1}$
C.
$24 \, ms^{-1}$
D.
$18 \, ms^{-1}$
Q377
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
Maximum value of velocity along positive x-direction is
A.
$35 \, ms^{-1}$
B.
$33 \, ms^{-1}$
C.
$27 \, ms^{-1}$
D.
$30 \, ms^{-1}$
Q378
Advance
LINKED COMPREHENSION TYPE QUESTIONS
MCQ
Comprehension Passage: Comprehension - 16
A particle is moving along x-axis and its initial velocity is $27 \, ms^{-1}$ . The acceleration of particle is given by the relation $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds. At t = 0 particle is at x = 0.
Based on the above facts, answer the following questions.
Maximum value of displacement along positive x-direction is
A.
54 m
B.
27 m
C.
120 m
D.
None of these
Q379
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
From the v-t graph shown in figure, match the quantities in COLUMN-I to their respective conclusions in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) between t = 0 and t = 1 s | (p) v = 0 |
| (B) between t = 1 s and t = 2 s | (q) a = 0 |
| (C) between t = 2 s and t = 3 s | (r) v ≠ 0 |
| (D) between t = 3 s and t = 4 s | (s) a ≠ 0 |
| (E) between t = 4 s and t = 5 s | (t) accelerated |
| (F) between t = 5 s and t = 6 s | (u) decelerated |
| (G) at t = 1 s and at t = 3 s |
Correct Answer: A → (s, u) B → (s, t) C → (s, u) D → (s, t) E → (q, r) F → (s, u) G → (p, s)
Explanation:
A → (s, u)
$\mathrm{B} \rightarrow (\mathrm{s}, \mathrm{t})$
$C \to (s, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, t)$
$\mathrm{E} \rightarrow (\mathrm{q}, \mathrm{r})$
$\mathrm{F} \rightarrow (\mathrm{s}, \mathrm{u})$
$G \to (p, s)$
Between t=0 and t=1 s, motion is decelerated $a\neq0$ .
$(\mathrm{A}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between t=1 s and t=2 s, motion is accelerated $a\neq0$ .
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between t = 2 s and t = 3 s, motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between $t = 3 \, \text{s}$ and $t = 4 \, \text{s}$ , motion is accelerated $a \neq 0$ .
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between $t = 4 \, \text{s}$ and $t = 5 \, \text{s}$ , motion is uniform i.e., $v \neq 0$ and $a = 0$ .
$\mathrm{So}, (\mathrm{E}) \rightarrow (\mathrm{q}, \mathrm{r})$
Between $t = 5 \, \text{s}$ and $t = 6 \, \text{s}$ , motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{F}) \rightarrow (\mathrm{s}, \mathrm{u})$
At t=1 s and at t=3 s the motion reverses its direction, so v=0 and simultaneously $a\neq0$ .
$\mathrm{So}, (\mathrm{G}) \rightarrow (\mathrm{p}, \mathrm{s})$
$\mathrm{B} \rightarrow (\mathrm{s}, \mathrm{t})$
$C \to (s, u)$
$\mathrm{D} \rightarrow (\mathrm{s}, t)$
$\mathrm{E} \rightarrow (\mathrm{q}, \mathrm{r})$
$\mathrm{F} \rightarrow (\mathrm{s}, \mathrm{u})$
$G \to (p, s)$
Between t=0 and t=1 s, motion is decelerated $a\neq0$ .
$(\mathrm{A}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between t=1 s and t=2 s, motion is accelerated $a\neq0$ .
$\mathrm{So}, (\mathrm{B}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between t = 2 s and t = 3 s, motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{C}) \rightarrow (\mathrm{s}, \mathrm{u})$
Between $t = 3 \, \text{s}$ and $t = 4 \, \text{s}$ , motion is accelerated $a \neq 0$ .
$\mathrm{So}, (\mathrm{D}) \rightarrow (\mathrm{s}, \mathrm{t})$
Between $t = 4 \, \text{s}$ and $t = 5 \, \text{s}$ , motion is uniform i.e., $v \neq 0$ and $a = 0$ .
$\mathrm{So}, (\mathrm{E}) \rightarrow (\mathrm{q}, \mathrm{r})$
Between $t = 5 \, \text{s}$ and $t = 6 \, \text{s}$ , motion is decelerated ( $a \neq 0$ ).
$\mathrm{So}, (\mathrm{F}) \rightarrow (\mathrm{s}, \mathrm{u})$
At t=1 s and at t=3 s the motion reverses its direction, so v=0 and simultaneously $a\neq0$ .
$\mathrm{So}, (\mathrm{G}) \rightarrow (\mathrm{p}, \mathrm{s})$
Q380
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For a particle moving rectilinearly, the x varies with t as per the equation $x = -5t^{2} + 20t + 10$ , where x is in metre and t is in second.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Average speed, in $ms^{-1}$ , from $t = 0$ to $t = 4$ s | (p) 20 |
| (B) Average velocity, in $ms^{-1}$ , from $t = 0$ to $t = 4$ s | (q) 10 |
| COLUMN-I | COLUMN-II |
|---|---|
| (C) Acceleration, in $ms^{-2}$ , at $t = 4$ s | (r) Zero |
| (D) Speed, in $ms^{-1}$ , at $t = 4$ s | (s) -4 |
| (t) None of these |
Correct Answer: A → (q) B → (r) C → (s) D → (p)
Explanation:
A → (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$x = - 5 t ^ {2} + 2 0 t + 1 0$
$\Rightarrow \frac {d x}{d t} = - 1 0 t + 2 0$
$\Rightarrow v = 2 0 - 1 0 t\tag{... (1}$
Now v=0 at t=2 s, so the distance travelled by the particle from t=0 to t=3 s is
$x = \left| \int_ {0} ^ {2} v d t \right| + \left| \int_ {2} ^ {3} v d t \right|$
$\mathrm{Now} \int v d t = (2 0 - 1 0 t) d t$
$\Rightarrow \int v d t = 2 0 t - 5 t ^ {2}$
$\Rightarrow x = \left| (2 0 t - 5 t ^ {2}) \right| _ {0} ^ {2} + \left| (2 0 t - 5 t ^ {2}) \right| _ {2} ^ {3}$
$\Rightarrow x = \vert 4 0 - 2 0 \vert + \vert 8 0 - 8 0 - 4 0 + 2 0 \vert$
$\Rightarrow x = | 2 0 | + | - 2 0 |$
$\Rightarrow x = 4 0 \mathrm{m}$
So, average speed is
$v _ {a v} = \frac {x}{t} = \frac {4 0}{4} = 1 0 \mathrm{ms} ^ {- 1}$
Displacement of the particle from t = 0 to t = 4 s is
$\Delta x = \int_ {0} ^ {4} v d t = \int_ {0} ^ {4} (2 0 - 1 0 t) d t$
$\Rightarrow \Delta x = (2 0 t - 5 t ^ {2}) \big | _ {0} ^ {4}$
$\Rightarrow \Delta x = 8 0 - 8 0 = 0$
So, average velocity is
$v _ {a v} = \frac {\Delta x}{\Delta t} = \frac {0}{4} = 0 \mathrm{ms} ^ {- 1}$
From (1), $a = \frac{dv}{dt} = -10\mathrm{ms}^{-2}$
$u = v | _ {t = 0} = 2 0 \mathrm{ms} ^ {- 1}$
Since, initially $u = \oplus$ and $a = \ominus$ , so a happens to be retardation for the motion from t = 0 to t = 2 s (till its velocity becomes zero momentarily i.e., the particle reverses its direction of motion).
Velocity at $t = 4$ s is
$v | _ {t = 4 \mathrm{s}} = 2 0 - 1 0 (4) = - 2 0 \mathrm{ms} ^ {- 1}$
Hence, speed at t = 4 s is $20 \, ms^{-1}$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (\mathrm{s})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$x = - 5 t ^ {2} + 2 0 t + 1 0$
$\Rightarrow \frac {d x}{d t} = - 1 0 t + 2 0$
$\Rightarrow v = 2 0 - 1 0 t\tag{... (1}$
Now v=0 at t=2 s, so the distance travelled by the particle from t=0 to t=3 s is
$x = \left| \int_ {0} ^ {2} v d t \right| + \left| \int_ {2} ^ {3} v d t \right|$
$\mathrm{Now} \int v d t = (2 0 - 1 0 t) d t$
$\Rightarrow \int v d t = 2 0 t - 5 t ^ {2}$
$\Rightarrow x = \left| (2 0 t - 5 t ^ {2}) \right| _ {0} ^ {2} + \left| (2 0 t - 5 t ^ {2}) \right| _ {2} ^ {3}$
$\Rightarrow x = \vert 4 0 - 2 0 \vert + \vert 8 0 - 8 0 - 4 0 + 2 0 \vert$
$\Rightarrow x = | 2 0 | + | - 2 0 |$
$\Rightarrow x = 4 0 \mathrm{m}$
So, average speed is
$v _ {a v} = \frac {x}{t} = \frac {4 0}{4} = 1 0 \mathrm{ms} ^ {- 1}$
Displacement of the particle from t = 0 to t = 4 s is
$\Delta x = \int_ {0} ^ {4} v d t = \int_ {0} ^ {4} (2 0 - 1 0 t) d t$
$\Rightarrow \Delta x = (2 0 t - 5 t ^ {2}) \big | _ {0} ^ {4}$
$\Rightarrow \Delta x = 8 0 - 8 0 = 0$
So, average velocity is
$v _ {a v} = \frac {\Delta x}{\Delta t} = \frac {0}{4} = 0 \mathrm{ms} ^ {- 1}$
From (1), $a = \frac{dv}{dt} = -10\mathrm{ms}^{-2}$
$u = v | _ {t = 0} = 2 0 \mathrm{ms} ^ {- 1}$
Since, initially $u = \oplus$ and $a = \ominus$ , so a happens to be retardation for the motion from t = 0 to t = 2 s (till its velocity becomes zero momentarily i.e., the particle reverses its direction of motion).
Velocity at $t = 4$ s is
$v | _ {t = 4 \mathrm{s}} = 2 0 - 1 0 (4) = - 2 0 \mathrm{ms} ^ {- 1}$
Hence, speed at t = 4 s is $20 \, ms^{-1}$
Q381
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the quantities in COLUMN-I with the corresponding expressions in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Velocity | (p) $\frac{d\vec{v}}{dt}$ |
| (B) Tangential acceleration | (q) $\frac{d\vec{r}}{dt}$ |
| (C) Acceleration | (r) $\frac{d|\vec{v}|}{dt}$ |
| (D) Instantaneous speed | (s) $\frac{d^{2}\vec{r}}{dt^{2}}$ |
| (t) $\left| \frac{d\vec{v}}{dt} \right|$ | |
| (u) None of these |
Correct Answer: A → (q) B → (r) C → (p, s) D → (t)
Explanation:
A $\rightarrow$ (q)
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (t)$
$\frac {d \vec {r}}{d t} = \mathrm{Velocity}$
$\frac {d | \vec {v} |}{d t} = \text { Tangential Acceleration }$
$\frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = \mathrm{Acceleration}$
$\left| \frac {d \vec {r}}{d t} \right| = \text { Instantaneous speed }$
$\mathrm{B} \rightarrow (\mathrm{r})$
$C \to (p, s)$
$\mathrm{D} \rightarrow (t)$
$\frac {d \vec {r}}{d t} = \mathrm{Velocity}$
$\frac {d | \vec {v} |}{d t} = \text { Tangential Acceleration }$
$\frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = \mathrm{Acceleration}$
$\left| \frac {d \vec {r}}{d t} \right| = \text { Instantaneous speed }$
Q382
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle moves such that its x coordinate is related to the time t by the relation $t = \sqrt{x} + 3$ , where x is in metre, t is in second. Based on this information, match the values in COLUMN-I (in SI units) to their respective quantities for the particles motion given in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) 0 | (p) Acceleration at $t = 5 \text{ s}$ . |
| (B) 2 | (q) Average speed from $t = 0$ to $t = 6 \text{ s}$ . |
| (C) 3 | (r) Velocity at the point of reversal of motion. |
| (D) 18 | (s) Total distance travelled from $t = 0$ to $t = 6 \text{ s}$ . |
| (t) Displacement from $t = 0$ to $t = 6 \text{ s}$ . |
Correct Answer: A → (r, t) B → (p) C → (q) D → (s)
Explanation:
A $\rightarrow (\mathbf{r},\mathbf{t})$
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$x = (t - 3) ^ {2}$
$\Rightarrow x = t ^ {2} - 6 t + 9$
$\Rightarrow v = \frac {d x}{d t} = 2 t - 6$
2 t - 6 = 0
$\Rightarrow t = 3 \mathrm{s}$
Displacement of the particle from t=0 to t=6 s is zero.
Distance travelled by the particle from t = 0 to t = 6 is $9 + 9 = 18 \, m$
Average speed of the particle is
$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {1 8}{6} = 3 \mathrm{ms} ^ {- 1}$
Average Velocity of the particle is
$\left| \vec {v} _ {a v} \right| = \frac {\text { Displacement }}{\text { Time }} = 0$
Acceleration of the particle over the entire duration of motion is $2 \, ms^{-2}$ .
$\mathrm{B} \rightarrow (\mathrm{p})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{s})$
$x = (t - 3) ^ {2}$
$\Rightarrow x = t ^ {2} - 6 t + 9$
$\Rightarrow v = \frac {d x}{d t} = 2 t - 6$
2 t - 6 = 0
$\Rightarrow t = 3 \mathrm{s}$
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| x | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
| v | -6 | -4 | -2 | 0 | 2 | 4 | 6 |
| a | 2 | 2 | 2 | 2 | 2 | 2 | 2 |
| Nature of Motion | Decelerating | Decelerating | Decelerating | Accelerating | Accelerating | Accelerating | Accelerating |
Displacement of the particle from t=0 to t=6 s is zero.
Distance travelled by the particle from t = 0 to t = 6 is $9 + 9 = 18 \, m$
Average speed of the particle is
$v _ {a v} = \frac {\text { Total Distance Travelled }}{\text { Total Time Taken }}$
$\Rightarrow v _ {a v} = \frac {1 8}{6} = 3 \mathrm{ms} ^ {- 1}$
Average Velocity of the particle is
$\left| \vec {v} _ {a v} \right| = \frac {\text { Displacement }}{\text { Time }} = 0$
Acceleration of the particle over the entire duration of motion is $2 \, ms^{-2}$ .
Q383
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A man can row a boat with $4 \, kmh^{-1}$ in still water. The man wishes to cross the river of width 4 km having a water current of $2 \, kmhr^{-1}$ . To cross the river with zero drift he swims making at an angle $\alpha$ degree with the current flow taking a time $t_{1}$ minutes to cross the river.
Now he wishes to cross the river in the shortest time $t_{2}$ minutes making an angle $\beta$ degree with the river flow.
Further he takes a time $t_{3}$ minutes to row 2 km upstream and then downstream back to the start point. Assuming all the cases to be independent of each other, the man to start from the river bank from the same point in the first two cases and from the midpoint of the river in the third case, match the quantities in COLUMN-I to the values in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) $\alpha$ | (p) $40\sqrt{3}$ |
| (B) $\beta$ | (q) 60 |
| (C) $t_{1}$ | (r) 80 |
| (D) $t_{2}$ | (s) 90 |
| (E) $t_{3}$ | (t) 120 |
| (u) Zero |
Correct Answer: A → (t) B → (s) C → (p) D → (q) E → (r)
Explanation:
A → (t)
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{E} \rightarrow (\mathrm{r})$
Given, that $v_{br} = 4 \, kmhr^{-1}$ and $v_{r} = 2 \, kmhr^{-1}$
$\Rightarrow \quad \alpha = \sin^ {- 1} \left(\frac {v _ {r}}{v _ {b r}}\right) = \sin^ {- 1} \left(\frac {2}{4}\right) = \sin^ {- 1} \left(\frac {1}{2}\right) = 3 0 ^ {\circ}$
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of $30^{\circ}$ with AB or $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the river flow.
Time taken by the boatman to cross the river for zero drift condition is
$t _ {1} = \frac {l}{v _ {b r} \cos \alpha}$
where $l = 4 \, km$ , $v_{br} = 4 \, kmh^{-1}$ and $\alpha = 30^{\circ}$
$\Rightarrow t _ {1} = \frac {4}{4 \cos (3 0 ^ {\circ})} = \frac {2}{\sqrt {3}} \mathrm{hr}$
$\Rightarrow t_{1} = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ minute
For shortest time $\theta=0^{\circ}$ i.e., the man must row the boat perpendicular to the river flow. So $\beta=90^{\circ}$
and $t_{\min}=t_{2}=\frac{l}{v_{br}\cos(0^{\circ})}=\frac{4}{4}=1\mathrm{hr}=60\mathrm{minute}$
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time $t_{2}$ of 60 minute.
$t = t _ {P \to Q} + t _ {Q \to P}$
$\Rightarrow t = \frac {P Q}{v _ {b r} + v _ {r}} + \frac {Q P}{v _ {b r} + v _ {r}}$
$\Rightarrow t = \frac {2}{4 + 2} + \frac {2}{4 - 2} = \frac {4}{3} \mathrm{hr}$
$\Rightarrow$ t = 80 minute
$\mathrm{B} \rightarrow (\mathrm{s})$
$C \rightarrow (p)$
$\mathrm{D} \rightarrow (\mathrm{q})$
$\mathrm{E} \rightarrow (\mathrm{r})$
Given, that $v_{br} = 4 \, kmhr^{-1}$ and $v_{r} = 2 \, kmhr^{-1}$
$\Rightarrow \quad \alpha = \sin^ {- 1} \left(\frac {v _ {r}}{v _ {b r}}\right) = \sin^ {- 1} \left(\frac {2}{4}\right) = \sin^ {- 1} \left(\frac {1}{2}\right) = 3 0 ^ {\circ}$
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of $30^{\circ}$ with AB or $90^{\circ} + 30^{\circ} = 120^{\circ}$ with the river flow.
Time taken by the boatman to cross the river for zero drift condition is
$t _ {1} = \frac {l}{v _ {b r} \cos \alpha}$
where $l = 4 \, km$ , $v_{br} = 4 \, kmh^{-1}$ and $\alpha = 30^{\circ}$
$\Rightarrow t _ {1} = \frac {4}{4 \cos (3 0 ^ {\circ})} = \frac {2}{\sqrt {3}} \mathrm{hr}$
$\Rightarrow t_{1} = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ minute
For shortest time $\theta=0^{\circ}$ i.e., the man must row the boat perpendicular to the river flow. So $\beta=90^{\circ}$
and $t_{\min}=t_{2}=\frac{l}{v_{br}\cos(0^{\circ})}=\frac{4}{4}=1\mathrm{hr}=60\mathrm{minute}$
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time and this shortest time $t_{2}$ of 60 minute.
$t = t _ {P \to Q} + t _ {Q \to P}$
$\Rightarrow t = \frac {P Q}{v _ {b r} + v _ {r}} + \frac {Q P}{v _ {b r} + v _ {r}}$
$\Rightarrow t = \frac {2}{4 + 2} + \frac {2}{4 - 2} = \frac {4}{3} \mathrm{hr}$
$\Rightarrow$ t = 80 minute
Q384
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For one dimensional motion if $v_{av}$ be the average speed, $\vec{v}_{av}$ be the average velocity, $v_{inst}$ be the instantaneous speed, $\vec{v}_{inst}$ be the instantaneous velocity and v be the speed, then match the following
| COLUMN-I | COLUMN-II |
|---|---|
| (A) $\vec{v}_{\text{inst}} = \vec{v}_{av}$ | (p) for uniform motion in any direction |
| (B) $|\vec{v}_{\text{inst}}| = v$ | (q) for uniform motion in given direction |
| (C) $v_{\text{inst}} = v_{av}$ | (r) Always true |
| (D) $|\vec{v}_{\text{inst}}| < v$ | (s) Never true |
Correct Answer: A → (q) B → (r) C → (p, q) D → (s)
Explanation:
A → (q) B → (r) C → (p, q) D → (s)
Q385
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Motion of dropped ball | (p) Two dimensional motion |
| (B) Motion of a snake | (q) Three dimensional motion |
| (C) Motion of a bird | (r) One dimensional motion |
| (D) Earth | (s) Absolute rest |
Correct Answer: A → (r) B → (p) C → (q) D → (p)
Explanation:
A → (r) B → (p) C → (q) D → (p)
Q386
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The displacement-time graph of a body moving on a straight line is given by
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Velocity – time graph | (p) |
| (B) Acceleration-time graph | (q) |
| COLUMN-I | COLUMN-II |
|---|---|
| (C) Distance – time graph | (r)  |
| (D) Speed – time graph | (s)  |
Correct Answer: A → (s) B → (r) C → (p) D → (q)
Explanation:
A → (s) B → (r) C → (p) D → (q)
Q387
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
For the velocity-time graph shown in figure, in a time interval from t=0 to t=6 s, match the following
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Change in velocity | (p) $-\frac{5}{3}$ SI unit |
| (B) Average acceleration | (q) $-20$ SI unit |
| (C) Total displacement | (r) $-10$ SI unit |
| (D) Acceleration at $t = 3$ s | (s) $-5$ SI unit |
Correct Answer: A → (r) B → (p) C → (r) D → (s)
Explanation:
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{r}) \\ \mathrm{D} \to (\mathrm{s}) \end{array}$
$v _ {i} = + 1 0 \mathrm{ms} ^ {- 1} \text {and} v _ {f} = 0$
$\Rightarrow \Delta v = v _ {f} - v _ {f} = - 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow a _ {a v} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = \frac {- 5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = area under v-t graph (with sign) and acceleration = slope of v-t graph.
$v _ {i} = + 1 0 \mathrm{ms} ^ {- 1} \text {and} v _ {f} = 0$
$\Rightarrow \Delta v = v _ {f} - v _ {f} = - 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow a _ {a v} = \frac {\Delta v}{\Delta t} = \frac {- 1 0}{6} = \frac {- 5}{3} \mathrm{ms} ^ {- 2}$
Total displacement = area under v-t graph (with sign) and acceleration = slope of v-t graph.
Q388
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A balloon rises up with constant net acceleration of $10 \, ms^{-2}$ . After 2 s a particle drops from the balloon. After further 2 s match the following (Take $g = 10 \, ms^{-2}$ )
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Height of particle ground | (p) Zero |
| (B) Speed of particle | (q) 10 SI units |
| (C) Displacement of particle | (r) 40 SI units |
| (D) Acceleration of particle | (s) 20 SI units |
Correct Answer: A → (r) B → (p) C → (s) D → (q)
Explanation:
A → (r)
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{s}) \\ \mathrm{D} \to (\mathrm{q}) \end{array}$
After 2 s velocity of balloon and hence the velocity of the particle will be $20 \, \text{ms}^{-1} (= \text{at})$ and its height from the ground will be $20 \, m \left( = \frac{1}{2} at^{2} \right)$ . Now g, will start acting on the particle.
$\begin{array}{l} \mathrm{A} \to (\mathrm{r}) \\ \mathrm{B} \to (\mathrm{p}) \\ \mathrm{C} \to (\mathrm{s}) \\ \mathrm{D} \to (\mathrm{q}) \end{array}$
After 2 s velocity of balloon and hence the velocity of the particle will be $20 \, \text{ms}^{-1} (= \text{at})$ and its height from the ground will be $20 \, m \left( = \frac{1}{2} at^{2} \right)$ . Now g, will start acting on the particle.
Q389
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A body accelerates from rest for time $t_{1}$ at a constant rate $\alpha$ for distance x then it decelerates at constant rate $\beta$ for time $t_{2}$ and covers distance y in this time and come at rest. If all quantities are in SI units, then match the following columns.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) $\frac{x}{y}$ | (p) $\frac{t_{1}}{t_{2}}$ |
| (B) $\frac{\alpha}{\beta}$ | (q) $\frac{t_{2}}{t_{1}}$ |
| (C) average speed for whole journey | (r) $\sqrt{\frac{2\alpha\beta}{\alpha+\beta}(x+y)}$ |
| (D) Maximum speed attained in it whole journey | (s) $\sqrt{\frac{\alpha\beta}{\alpha+\beta}\left(\frac{x+y}{2}\right)}$ |
Correct Answer: A → (p) B → (q) C → (s) D → (r)
Explanation:
A → (p)
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \to (s)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$\frac {v _ {m}}{t _ {1}} = \alpha \mathrm{and} \frac {v _ {m}}{t _ {2}} = \beta$
$\text { Also } x + y = \frac {1}{2} \times v _ {m} \times \left(t _ {1} + t _ {2}\right)$
$\Rightarrow \quad (x + y) = \frac {1}{2} \times v _ {m} \times \left(\frac {v _ {m}}{\alpha} + \frac {v _ {m}}{\beta}\right)$
$\Rightarrow v _ {m} = \sqrt {\frac {2 (x + y) \alpha \beta}{(\alpha + \beta)}}$
Now, $v_{av}=\frac{x+y}{t_{1}+t_{2}}=\frac{x+y}{\frac{v_{m}}{\alpha}+\frac{v_{m}}{\beta}}=\frac{\alpha\beta}{\alpha+\beta}\cdot\frac{x+y}{v_{m}}$
$\mathrm{B} \rightarrow (\mathrm{q})$
$C \to (s)$
$\mathrm{D} \rightarrow (\mathrm{r})$
$\frac {v _ {m}}{t _ {1}} = \alpha \mathrm{and} \frac {v _ {m}}{t _ {2}} = \beta$
$\text { Also } x + y = \frac {1}{2} \times v _ {m} \times \left(t _ {1} + t _ {2}\right)$
$\Rightarrow \quad (x + y) = \frac {1}{2} \times v _ {m} \times \left(\frac {v _ {m}}{\alpha} + \frac {v _ {m}}{\beta}\right)$
$\Rightarrow v _ {m} = \sqrt {\frac {2 (x + y) \alpha \beta}{(\alpha + \beta)}}$
Now, $v_{av}=\frac{x+y}{t_{1}+t_{2}}=\frac{x+y}{\frac{v_{m}}{\alpha}+\frac{v_{m}}{\beta}}=\frac{\alpha\beta}{\alpha+\beta}\cdot\frac{x+y}{v_{m}}$
Q390
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The equation of one dimensional motion of particle is described in COLUMN-I. At t = 0, particle is at origin and at rest. Match the COLUMN-I with the statements in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) $x = (3t^{2} + 2) \text{ m}$ | (p) velocity of particle at $t = 1 \text{ s is } 8 \text{ ms}^{-1}$ |
| (B) $v = 8t \text{ ms}^{-1}$ | (q) particle moves with uniform acceleration |
| (C) $a = 16t$ | (r) particle moves with variable acceleration |
| (D) $v = 6t - 3t^{2}$ | (s) particle will change its direction some time |
Correct Answer: A → (q) B → (p, q) C → (p, r) D → (r, s)
Explanation:
A → (q)
B → (p, q)
C → (p, r)
D → (r, s) $v = \frac{dx}{dt}$ $\Rightarrow a = \frac{dv}{dt}$ $\Rightarrow v = \int_{0}^{t} ad t$ $\Rightarrow x = \int_{0}^{t} v dt$
B → (p, q)
C → (p, r)
D → (r, s) $v = \frac{dx}{dt}$ $\Rightarrow a = \frac{dv}{dt}$ $\Rightarrow v = \int_{0}^{t} ad t$ $\Rightarrow x = \int_{0}^{t} v dt$
Q391
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
v-t graph of a particle moving along positive direction x is shown in figure. Match the items in COLUMN-I with the respective answers in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) a-x graph | (p) Parabola |
| COLUMN-I | COLUMN-II |
|---|---|
| (B) v-x graph | (q) Circle |
| (C) a-t graph | (r) Straight line |
| (D) a-v graph | (s) Ellipse |
Correct Answer: A → (r) B → (p) C → (r) D → (r)
Explanation:
A → (r) B → (p) C → (r) D → (r)
Q392
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the v-t graphs in COLUMN-I with the respective a-t graphs in COLUMN-II.
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A → (r) B → (q) C → (s) D → (p)
Explanation:
A → (r)
B → (q)
C → (s)
D → (p)
A → v = Kt
$⇒ a = \frac{dv}{dt} = K$
$B → v = Kt^2$
$⇒ a = \frac{dv}{dt} = 2Kt$
$C → K_1t → First part of graph$
$⇒ a = \frac{dv}{dt} = K_1$
$⇒ v = K_2 → Second part of graph$
$⇒ a = \frac{dv}{dt} = 0$
$D → v = v_0e^{-Kt}$
$⇒ a = \frac{dv}{dt} = (-v_0K)e^{-Kt}$
$⇒ a = -a_0e^{-Kt}$
$⇒ a_0 = v_0K$
B → (q)
C → (s)
D → (p)
A → v = Kt
$⇒ a = \frac{dv}{dt} = K$
$B → v = Kt^2$
$⇒ a = \frac{dv}{dt} = 2Kt$
$C → K_1t → First part of graph$
$⇒ a = \frac{dv}{dt} = K_1$
$⇒ v = K_2 → Second part of graph$
$⇒ a = \frac{dv}{dt} = 0$
$D → v = v_0e^{-Kt}$
$⇒ a = \frac{dv}{dt} = (-v_0K)e^{-Kt}$
$⇒ a = -a_0e^{-Kt}$
$⇒ a_0 = v_0K$
Q393
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Let us call a motion, A when velocity is positive and increasing. $A^{-1}$ when velocity is negative and increasing. R when velocity is positive and decreasing and $R^{-1}$ when velocity is negative and decreasing. Now match the following two tables for the given s-t graph
| COLUMN-I | COLUMN-II |
|---|---|
| (A) M | (p) $A^{-1}$ |
| (B) N | (q) $R^{-1}$ |
| (C) P | (r) A |
| (D) Q | (s) R |
Correct Answer: A → (r) B → (s) C → (p) D → (q)
Explanation:
A → (r)
B → (s)
C → (p)
D → (q)
In motion M : slope of s-t graph is positive and increasing. Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
B → (s)
C → (p)
D → (q)
In motion M : slope of s-t graph is positive and increasing. Therefore, velocity of the particle is positive and increasing. Hence, it is A type motion. Similarly, N, P and Q can be observed from the slope.
Q394
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
In the $s-t$ equation $(s=10+20t-5t^{2})$ match the following
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Distance travelled in 3 s | (p) -20 unit |
| (B) Displacement in 1 s | (q) 15 unit |
| (C) Initial acceleration | (r) 25 unit |
| (D) Velocity at 4 s | (s) -10 unit |
Correct Answer: A → (r) B → (q) C → (s) D → (p)
Explanation:
A → (r) B → (q) C → (s) D → (p)
Q395
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The velocity time graphs for a particle moving along a straight line is given in each situation of COLUMN-I. Match the graph in COLUMN-I with corresponding statements in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
(A) | (p) Speed of particle is continuously decreasing. |
(B) | (q) Magnitude of acceleration of particle is decreasing with time. |
(C) | (r) Direction of acceleration of particle does not change. |
(D) | (s) Magnitude of acceleration of particle does not change. |
| (t) Acceleration is always opposite to the direction of velocity. |
Correct Answer: A → (r, s) B → (r, s) C → (p, q, r, t) D → (p, q, r, t)
Explanation:
A → (r, s) B → (r, s) C → (p, q, r, t) D → (p, q, r, t)
Q396
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the statements in COLUMN-I with corresponding graphs in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Particle moving with constant speed. | (p) |
| (B) Particle moving with increasing acceleration. |  |
| (C) Particle moving with constant negative acceleration. | (r) |
| (D) Particle moving with zero acceleration. | (s) |
Correct Answer: A → (q, s) B → (r) C → (p) D → (s)
Explanation:
A → (q, s) B → (r) C → (p) D → (s)
Q397
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
The motion of an object over time can often be communicated by graphs of its distance, velocity or acceleration with time. Different features of these graphs correspond to quantities of the motion. Match each quantity in the COLUMN-I with its graphical manifestation in the COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Distance travelled $\Delta d$ | (p) Slope of a distance-time graph |
| (B) Velocity change $\Delta v$ | (q) Slope of velocity-time graph |
| (C) Velocity $v$ | (r) Area under a velocity-time graph |
| (D) Acceleration $a$ | (s) Area under an acceleration-time graph. |
Correct Answer: A → (r) B → (s) C → (p) D → (q)
Explanation:
A → (r) B → (s) C → (p) D → (q)
Q398
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle is dropped vertically downward under gravity. Consider the downward direction as positive and the collision of the ball with the ground to be elastic, match the statements in COLUMN-I with corresponding graphs in COLUMN-II.
| COLUMN-I | COLUMN-II |
|---|---|
| (A) The distance travelled by particle varies with time as | (p) |
| (B) Velocity of particle changes with time as | (q) |
| (C) Displacement of particle depends on time as | (r) |
| (D) Dependency of acceleration on time is given by | (s) |
Correct Answer: A → (s) B → (r) C → (q) D → (p)
Explanation:
A → (s)
$\mathrm{C} \rightarrow (\mathrm{q})$
Acceleration is constant and is equal to acceleration due to gravity, which is acting vertically downwards i.e. in the positive direction, so (p) represents the a-t graph for the motion.
Since $v = u + at$ and $s = \Delta y = ut + \frac{1}{2}gt^{2}$
Before collision, v = gt {∵ u = 0}
So, before collision velocity is increasing linearly with time and is increasing. However, after collision velocity decreases with time linearly and is negative as in (r).
$y = \frac {1}{2} g t ^ {2}$
tive, whereas after collision, the displacement is $y = -v_{0}t + \frac{1}{2}gt^{2}$ and is positive (a parabola opening
(upwards). Here is the tricky part, as we have taken the origin at the point from where the ball is dropped and in both the cases of upward and downward motion, the displacement is downwards and hence displacement is positive in both the cases.
The distance-time graph is positive and always increasing.
$\mathrm{C} \rightarrow (\mathrm{q})$
Acceleration is constant and is equal to acceleration due to gravity, which is acting vertically downwards i.e. in the positive direction, so (p) represents the a-t graph for the motion.
Since $v = u + at$ and $s = \Delta y = ut + \frac{1}{2}gt^{2}$
Before collision, v = gt {∵ u = 0}
So, before collision velocity is increasing linearly with time and is increasing. However, after collision velocity decreases with time linearly and is negative as in (r).
$y = \frac {1}{2} g t ^ {2}$
tive, whereas after collision, the displacement is $y = -v_{0}t + \frac{1}{2}gt^{2}$ and is positive (a parabola opening
(upwards). Here is the tricky part, as we have taken the origin at the point from where the ball is dropped and in both the cases of upward and downward motion, the displacement is downwards and hence displacement is positive in both the cases.
The distance-time graph is positive and always increasing.
Q399
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
A particle is moving along $x$ -direction in four ways. Different graphs is plotted in COLUMN-I.
| COLUMN-I | COLUMN-II |
|---|---|
(A)  | (p) Variable velocity |
(B)  | (q) Positive acceleration |
(C)  | (r) Negative acceleration |
| (D) | (s) Constant speed |
 |
Correct Answer: A → (p, r) B → (p, r) C → (p, q) D → (p, q)
Explanation:
A → (p, r) B → (p, r) C → (p, q) D → (p, q)
Q400
Advance
MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS
Match the Columns
Match the following
| COLUMN-I | COLUMN-II |
|---|---|
| (A) Constant positive acceleration | (p) Speed may increase |
| (B) Constant negative acceleration | (q) Speed may decrease |
| (C) Constant displacement | (r) Speed is zero |
| (D) Constant slope of a-t graph | (s) Speed must increase |
Correct Answer: A → (p, q) B → (p, q) C → (r) D → (p, q)
Explanation:
A → (p, q)
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{q})$
With constant positive acceleration, speed increases, when velocity is positive and speed decreases, when velocity is negative.
Similarly, with constant negative acceleration speed increases, when velocity is negative and speed decreases, when velocity is positive.
$\mathrm{B} \rightarrow (\mathrm{p}, \mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{p}, \mathrm{q})$
With constant positive acceleration, speed increases, when velocity is positive and speed decreases, when velocity is negative.
Similarly, with constant negative acceleration speed increases, when velocity is negative and speed decreases, when velocity is positive.